Probability that a Randomly Chosen Vector is a Positive Linear Combination Choose a basis $\beta=\{v_i \in \mathbb{R}^n, 1\leq i\leq n\}$. What is the probability that a randomly chosen nonzero vector $x\in S^{n-1}\subset \mathbb{R}^n$ is a positive linear combination of vectors in $\beta$?
Note that, intuitively, the answer should depend on the vectors $v_i$ themselves.
 A: This is a nice question!! A partial answer to your question is:
If $\beta=\{v_1,v_2,...v_n\}$ is a basis for $\mathbb{R^{n}}$ then we can scale (by a positive number) the basis elements so that they sit on our $S^{n-1}$,Call the new basis vectors with the same name, now the area(or the '$n-1$' dimensional Volume) that could be reached by linear combination by positive scalars is the part of $S^{n-1}$ that is confined by the basis vectors ,so the probability is:
=$n-1$ volume of the confined region (boundaries will be the great sphere $S^{n-2}$) divided by Surface area of $S^{n-1}$
It can be quite difficult to solve this integral in general but we can always get a inequality for the bound as follows :
The $n-1$ area on the sphere reachable by positive coefficients are less than the area formed by the $n-1$ parallelopiped formed by these n points on $S^{n-1}$,so the probability is greater than or equal to (because the sphere is Convex) $\frac{{\sqrt{det(X.X^{T}})}/{2}}{V_S(S^{n-1})}$ where the Matrix $X$ is one with columns $v_{2}-v_1$ ,$v_3-v_1$ ,  ... ,$v_n-v_1$ 
